First slide

Electrical Resonance series circuit

Question

What is the value of inductance L for which the current is maximum in a series LCR circuit with $C = 10 μF$ and $ω = 1000 s^{- 1} ?$

Easy

A

100 mH
B

1 mH
C

Cannot be calculated unless R is known
D

10 mH

Solution

In resonance condition, maximum current flows in the circuit.
Current in LCR series circuit,
$i = \frac{V}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}$
Where V is rms value of current, R is resistance, X_L is inductive reactance and X_C is capacitive reactance. For current to be maximum,
$X_{L} = X_{C}$
This happens in resonance state of the circuit, i.e.,
$ωL = \frac{1}{ωC}$
or $L = \frac{1}{ω^{2} C}$ ….(i)
Given, $ω = 1000 s^{- 1}, C = 10 μF = 10 \times 10^{- 6} F$
Hence, $L = \frac{1}{(1000)^{2} \times 10 \times 10^{- 6}} = 0 .1 H = 100 mH$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App